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Section 5.1
123
(d) rng
f
=
{n
E
Z
I
n
~
998}.
26. [BB]
(a)
i. 0
E
nZ
because 0
=
O(n)
is a multiple of
n.
ii.
If
a
E
nZ,
then
a
=
kn
for some
k,
so
a
= (
k)n
E
nZ
because it is also a multiple of
n.
iii.
If
a, bEnZ,
then
a
=
kIn
and
b
=
k2n
for some integers
kl' k2'
so
a+ b
=
(kl +k2)n
E
nZ
because it is also a multiple of
n.
(b)
i. 0
E
A
by definition of ideal, so if
A
contains just one element, that element must be 0, in
which case
A
=
OZ is of the form
nZ.
ii.
If
A
contains more than element, it contains a nonzero element
a.
By definition of ideal,
A
contains both
a
and
a,
one of which is positive.
iii. The set of positive integers in
A
is not empty (by ii.) and, hence, contains a smallest element
by the WellOrdering Principle (4.1.2).
iv. First,
In
=
n
E
A.
Then, if
kn
E
A
for some
k
> 0, so also is
(k
+
l)n
E
A,
by definition,
because
(k
+
l)n
=
kn
+
n.
By the Principle of Mathematical Induction, we conclude that
kn
E
A
for all
k
>
O. Since
(k)n
=
kn
is the negative of
kn,
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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